Question

A body cools according to Newton’s law from \[{{100}^{\circ }}C\] to \[{{60}^{\circ }}C\] in 20 minutes. The temperature of the surrounding being \[{{20}^{\circ }}C\]. How long will it take to cool down to \[{{30}^{\circ }}C\].

Answer 1

Hint: In this problem, we are given a body cools according to Newton’s law from \[{{100}^{\circ }}C\] to \[{{60}^{\circ }}C\] in 20 minutes. The temperature of the surrounding being \[{{20}^{\circ }}C\]. We should find the time taken by the body to cool down to \[{{30}^{\circ }}C\]. We can first write the equation of Newton’s law of cooling. We can then integrate them and substitute the required data to find the time.

Complete step by step answer:
 Here we are given a body that cools according to Newton’s law from \[{{100}^{\circ }}C\] to \[{{60}^{\circ }}C\] in 20 minutes. The temperature of the surrounding being \[{{20}^{\circ }}C\]. We should find the time taken by the body to cool down to \[{{30}^{\circ }}C\].
Let \[{{\theta }^{\circ }}C\] be the temperature of the body at time t.
We know that according to Newton’s law of cooling, we can write the equation as,
\[\begin{align}
  & \Rightarrow \dfrac{d\theta }{dt}\propto \left( \theta -20 \right) \\
 & \Rightarrow \dfrac{d\theta }{dt}=-k\left( \theta -20 \right),k>0 \\
\end{align}\]
We can now write the above step as,
\[\Rightarrow \dfrac{d\theta }{\left( \theta -20 \right)}=-kdt\]
We can now integrate on both sides, we get
\[\begin{align}
  & \Rightarrow \int{\dfrac{d\theta }{\left( \theta -20 \right)}}=-k\int{dt} \\
 & \Rightarrow \log \left( \theta -20 \right)=-kt+C.......(1) \\
\end{align}\]
We know that initially we have t = 0, \[\theta ={{100}^{\circ }}C\], substituting in (1), we get
\[\begin{align}
  & \Rightarrow \log \left( 100-20 \right)=-k\left( 0 \right)+C \\
 & \Rightarrow C=\log 80......(2) \\
\end{align}\]
We can substitute (2) in (1), we get
\[\begin{align}
  & \Rightarrow \log \left( \theta -20 \right)=-kt+\log 80 \\
 & \Rightarrow \log \left( \dfrac{\theta -20}{80} \right)=-kt \\
\end{align}\]
Now, when t = 20 and \[\theta ={{60}^{\circ }}\], substituting in the above step, we get
\[\begin{align}
  & \Rightarrow \log \left( \dfrac{60-20}{80} \right)=-k\times 20 \\
 & \Rightarrow k=-\dfrac{1}{20}\log \left( \dfrac{1}{2} \right).......(3) \\
\end{align}\]
 Now (1) becomes,
\[\Rightarrow \log \left( \dfrac{\theta -20}{80} \right)=\dfrac{t}{20}\log \left( \dfrac{1}{2} \right)\]
We can now substitute \[\theta ={{30}^{\circ }}\] in the above step, we get
 \[\begin{align}
  & \Rightarrow \log \left( \dfrac{30-20}{80} \right)=\dfrac{t}{20}\log \left( \dfrac{1}{2} \right) \\
 & \Rightarrow \log \left( \dfrac{1}{8} \right)=\log {{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{20}}} \\
\end{align}\]
We can now write the above step as,
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{20}}} \\
 & \Rightarrow 3=\dfrac{t}{20} \\
 & \Rightarrow t=60\min \\
\end{align}\]
Therefore, the required time is 60 minutes.

Note: We should always remember that the Newton’s law of cooling equation is \[\dfrac{d\theta }{dt}\propto \left( \theta -20 \right)\], where we can take the proportion and add a constant equating the both sides. We should also know some of the integration formulas to solve these types of problems.